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Artillery Simulation

  1. Aug 4, 2004 #1
    I was wondering if yall could help me out with creating some formulas to use in an artillery simulation. I'm barly in HS and this sort of thing blows me out of the water.

    Theres 3 parts you need, I have the first mostly figured out.

    You are given the latitude and longitude of both your position and the the targets, there is no wind.

    I figured out how to find distance to the target, the first part.

    T1 is targets latitude
    T2 Is targets longitude
    A1 is the artillery peice lat
    A2 is the artillery peice long
    X is the difference in Lat
    Y is the difference in long
    D is the total Difference
    R is the end range

    T1-A1=X
    T2-A2=Y
    X+Y=D
    D/2=R

    The last thing I need to do is convert R (say 6.7'.32'') into meteres if at all possible. I can't find a formula anywhere.

    The second part is finding the degree I must aim left/right. 0 deg is directlly N, 90 is directly W 180 S ect. Is there a way to convert slope into degrees? And if so how would I find the slope of a line to my target?

    The third and least important part is the degree I must aim upwards to hit a target at X range. EDIT: There is an internal range adjuster in the sights that can be set up to 5km

    A bunch of info on the artillery I'll be using:

    Initial Projectile Speed:
    HE - 820m/s (High-Explosive)
    AP - 795m/s (Armor-Piercing)

    Projectile Weight:
    HE - 9kg
    AP - 9.5kg

    Barrel Length: 4.93m
    Barrel Elevation off ground: 6 ft

    P.S. If the difference in elevation was no larger than 25m, how bad would that throw my aim?

    I'd greatly appreciate any help at all, ar at least how I can figure it out myself?
     
  2. jcsd
  3. Aug 4, 2004 #2

    NateTG

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    In Practice this is *hard*

    Not that you want to hear this, but even relatively simple looking artillery problems can show up, and be daunting, in higher level math classes.

    Is this some sort of project of yours, or something that was assigned by a teacher? What you want to is a bit hairy unless there are a large number of simplifying assumptions.

    Regarding longditude and lattitute distance measurements:
    You can't add or subtract them meaningfully. If you like, take a look at a globe, and see what happens to lattitude and longditude near the north pole, and you'll see that it can get hairy. You should also be able to see why there isn't a formula for converting latitude/longditude angle to distance.

    So, let's make some additional starting assumptions to simplify things:
    The earth is spherical.
    The earth does not rotate.
    The path that the projectile should travel is 'along' the shortest surface line between the two points.
    The acceleration of gravity is not affected by altitude.

    Now, to determine the orientation that the gun needs to be pointed in, you need to find the great circle that cointains the gun and the target. Calculating the elevation to fire the gun at is not easy since the usual parabolic flight path is only valid for short distances, and may involve some pretty tricky calculus.

    To see how hard this type of problem is, with a globe in front of you try to figure out what angle would be necessary to fire a projectile from the north pole to the south pole.

    In real life, this type of problem is a bit harder. Once again, using a globe, imagine firing a projectile from Bhagdad to London and account for air resistance and the rotation of the earth. For bonus points, imagine firing cannons from a boat in the ocean where it's being rocked by the waves.
     
  4. Aug 4, 2004 #3
    Thnaks for the reply, this is a project of mine.

    Two more questions, and I'll leave ya alone :)

    1. What information would I need to find the great circle that contains the target and the artillery?

    2. The area that I'd be in woulden't be longer than 30km N-S, and 80km EW. Located around holland. Would the difference be small enough for me to measure the size of 1'' and then be able to determine distance?

    If not, could I use the rate of change in the angle of latitude and logitude from the equator? Well nevermind that sounds way out there.

    What other information could I use to make it easier?

    simplifying assumptions:

    The earth is flat.
    The earth does not rotate.
    The path that the projectile should travel is 'along' the shortest surface line between the two points.
    The acceleration of gravity is not affected by altitude.
    There are no wind currents.
    The target is 5m by 5m.
    The cannon is not affected by dispersion.
    An unlimited number of rounds can be used.

    Does that help any?
     
  5. Aug 4, 2004 #4

    NateTG

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    Ok, I recently suggested something like what you initally described as a sadistic physics problem.

    If you assume that the earth is flat, you don't need to worry about that, you can just use a 'straight line'.
    So lattititude and longditude are really not so big.

    Depending on how precise you want to be, you can either determine distance per degree (you will probably want to do lattitude and longditude seperately) or you can do something fancier. If you do that, you will need to use the pythagorean formula:
    [tex]d=\sqrt{a^2+b^2}[/tex]

    Then to find the distance that the projectile would travel on a flat surface, you can use the formula:
    [tex]d=\frac{v^2 \sin 2\theta}{2g}[/tex]
    Where
    [tex]d[/tex] is the distance you want the projectile to go
    [tex]v[/tex] is the inital velocity of the projectile
    and
    [tex]g[/tex] is the acceleration of gravity (about 9.81 m/s).

    PS:
    The muzzle velocities you give are about 1 km/sec. Let's say that you're shooting at a target 40 km away so the projectile will be in the air for at least 40 seconds. In that time, the earth will have rotated about 10m, so you will not be able to hit a 5x5meter target without accounting for the rotation.
     
    Last edited: Aug 4, 2004
  6. Aug 4, 2004 #5
    Thanks again Nate,

    So when I find the distance, how would I find the degree horizontally? Slope? How would I find the slope of the line?
     
  7. Aug 4, 2004 #6

    NateTG

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    I had transferred the formula incorrectly, it should be:
    [tex] \frac{v^2 \sin (2\theta)}{2g}[/tex]
    You can solve that for angle:
    [tex]d=\frac{v^2 \sin (2\theta)}{2g}[/tex]
    so
    [tex]\theta=\frac{\sin^{-1}\frac{2dg}{v^2}}{2}[/tex]
     
  8. Aug 4, 2004 #7
    I'm a little lost again, is (20) the angle of traverse you would shoot for?

    So to find distance would be:
    T1-A1=X
    T2-A2=Y
    Xdeg = Xmeters
    Ydeg = Ymeters
    X+Y=D
    D/2=R

    Angle of elevation:
    [tex]d=\frac{v^2 \sin (2\theta)}{2g}[/tex]

    Where 20 is the degree upwards

    Angle of traverse:
    [tex]\theta=\frac{\sin^{-1}\frac{2dg}{v^2}}{2}[/tex]

    Where 0 is the degree horizontaly?
     
  9. Aug 4, 2004 #8

    NateTG

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    No, θ is the angle off the horizontal, and you're still not calculating the distance correctly.

    Have fun shelling the Danes.
     
    Last edited: Aug 4, 2004
  10. Aug 4, 2004 #9
    Ok so If I was to fire at a point 1.1.1N by 1.1.1E from point 1.3.2N by 1.3.2E where 0.0.1 lat and long was equal to 10m

    The first step is to find distance to target, with a 1.210 km difference for both latitude and longitude, added togeather and divided by 2 equals 1210.

    Thats not correct?

    Seconed, angle of traverse:

    [tex]d=\frac{v^2 \sin (2\theta)}{2g}[/tex]

    I don't think Iv'e gotten that right, or at least how to compute it. (I have a graphing calculator)

    Angle off horizon:

    [tex]\theta=\frac{\sin^{-1}\frac{2dg}{v^2}}{2}[/tex]

    Same way with this one.

    Sorry to be a pain :\
     
  11. Aug 5, 2004 #10
    Ok so I need 3 things: The first is distance, the seconed is the angle of traverse (Directlly N being 0 deg), and the third is angle off the horizon.

    simplifying assumptions:

    The earth is flat.
    The earth does not rotate.
    The path that the projectile should travel is 'along' the shortest surface line between the two points.
    The acceleration of gravity is not affected by altitude.
    There are no wind currents.
    The target is 5m by 5m.
    The cannon is not affected by dispersion.
    An unlimited number of rounds can be used.
    The earth exists in a vacumme

    The distance formula works without a hitch so far, 00.00.01 is equal to 1.6667m for both latitude and longitude. So to convert it to m I take the difference, say 0.4.53 by first multiplying the number with the highest value (0 in this case) and adding that number to the seconed. Then I multiply that and add it to the last, and lastly multiply that by 1.6667m.

    Using a protractor against a grid I can somewhat guesstimate the angle of traverse, but I know that a protractor uses circumfrance of a circle to find the angle. So is it possible to use a "protractor formula" that instead of using visable lines, computes the answer using lines in numerical form, like slope? So it would be somthing along, the line of slope X intersects line of slope Y, at point A and could then find the answer?

    As for the final part, I'm not sure I understand the formulas you gave me, could you provide a link to a site explaining it or somthing along those lines?

    In the sights of the gun there is a range adjuster that goes up to 5km, but often It's hard put the crosshairs on where the target would be if you could see it, the angles are so low to the ground objects could get in the way, and isn't usefull above 5k

    Finally aren't there two possible angles for each distance, one low and one high? And if so would simply flipping the value along 45 deg work? 40 deg would be 50 deg, 30 would be 60, ect.ect.

    I'm shelling the gerrys :)

    Thanks again for all the help, I'm learning alot more than I expected.
     
  12. Aug 5, 2004 #11

    NateTG

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    As a representative of the boche I must object.
    Of course, at the rate you're going you'll have a hard time hitting anything ;)

    If the target is [tex]x[/tex] km to the North, and [tex]y[/tex] km to the West, then the angle (going West from North) to the target is one of the two solutions to [tex]\phi=\tan^{-1}(\frac{y}{x})[/tex].

    The distance to the target is [tex]\sqrt{x^2+y^2}[/tex].

    Those two will make a lot more sense if you look into bit of trignometry and Pythagoras' theorem.

    The angle you want to shoot above horizontal is [tex]\frac{\sin^{-1}\frac{2g\sqrt{x^2+y^2}}{v^2}}{2}[/tex]
    Which will have multiple answers some of which don't make sense.
     
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